| 1 | /*
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| 2 | * Copyright 2000-2021 The OpenSSL Project Authors. All Rights Reserved.
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| 3 | *
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| 4 | * Licensed under the Apache License 2.0 (the "License"). You may not use
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| 5 | * this file except in compliance with the License. You can obtain a copy
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| 6 | * in the file LICENSE in the source distribution or at
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| 7 | * https://www.openssl.org/source/license.html
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| 8 | */
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| 9 |
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| 10 | #include "internal/cryptlib.h"
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| 11 | #include "bn_local.h"
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| 12 |
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| 13 | BIGNUM *BN_mod_sqrt(BIGNUM *in, const BIGNUM *a, const BIGNUM *p, BN_CTX *ctx)
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| 14 | /*
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| 15 | * Returns 'ret' such that ret^2 == a (mod p), using the Tonelli/Shanks
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| 16 | * algorithm (cf. Henri Cohen, "A Course in Algebraic Computational Number
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| 17 | * Theory", algorithm 1.5.1). 'p' must be prime!
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| 18 | */
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| 19 | {
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| 20 | BIGNUM *ret = in;
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| 21 | int err = 1;
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| 22 | int r;
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| 23 | BIGNUM *A, *b, *q, *t, *x, *y;
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| 24 | int e, i, j;
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| 25 | int used_ctx = 0;
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| 26 |
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| 27 | if (!BN_is_odd(p) || BN_abs_is_word(p, 1)) {
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| 28 | if (BN_abs_is_word(p, 2)) {
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| 29 | if (ret == NULL)
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| 30 | ret = BN_new();
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| 31 | if (ret == NULL)
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| 32 | goto end;
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| 33 | if (!BN_set_word(ret, BN_is_bit_set(a, 0))) {
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| 34 | if (ret != in)
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| 35 | BN_free(ret);
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| 36 | return NULL;
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| 37 | }
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| 38 | bn_check_top(ret);
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| 39 | return ret;
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| 40 | }
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| 41 |
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| 42 | ERR_raise(ERR_LIB_BN, BN_R_P_IS_NOT_PRIME);
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| 43 | return NULL;
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| 44 | }
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| 45 |
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| 46 | if (BN_is_zero(a) || BN_is_one(a)) {
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| 47 | if (ret == NULL)
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| 48 | ret = BN_new();
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| 49 | if (ret == NULL)
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| 50 | goto end;
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| 51 | if (!BN_set_word(ret, BN_is_one(a))) {
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| 52 | if (ret != in)
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| 53 | BN_free(ret);
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| 54 | return NULL;
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| 55 | }
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| 56 | bn_check_top(ret);
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| 57 | return ret;
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| 58 | }
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| 59 |
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| 60 | BN_CTX_start(ctx);
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| 61 | used_ctx = 1;
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| 62 | A = BN_CTX_get(ctx);
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| 63 | b = BN_CTX_get(ctx);
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| 64 | q = BN_CTX_get(ctx);
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| 65 | t = BN_CTX_get(ctx);
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| 66 | x = BN_CTX_get(ctx);
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| 67 | y = BN_CTX_get(ctx);
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| 68 | if (y == NULL)
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| 69 | goto end;
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| 70 |
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| 71 | if (ret == NULL)
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| 72 | ret = BN_new();
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| 73 | if (ret == NULL)
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| 74 | goto end;
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| 75 |
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| 76 | /* A = a mod p */
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| 77 | if (!BN_nnmod(A, a, p, ctx))
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| 78 | goto end;
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| 79 |
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| 80 | /* now write |p| - 1 as 2^e*q where q is odd */
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| 81 | e = 1;
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| 82 | while (!BN_is_bit_set(p, e))
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| 83 | e++;
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| 84 | /* we'll set q later (if needed) */
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| 85 |
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| 86 | if (e == 1) {
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| 87 | /*-
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| 88 | * The easy case: (|p|-1)/2 is odd, so 2 has an inverse
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| 89 | * modulo (|p|-1)/2, and square roots can be computed
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| 90 | * directly by modular exponentiation.
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| 91 | * We have
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| 92 | * 2 * (|p|+1)/4 == 1 (mod (|p|-1)/2),
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| 93 | * so we can use exponent (|p|+1)/4, i.e. (|p|-3)/4 + 1.
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| 94 | */
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| 95 | if (!BN_rshift(q, p, 2))
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| 96 | goto end;
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| 97 | q->neg = 0;
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| 98 | if (!BN_add_word(q, 1))
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| 99 | goto end;
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| 100 | if (!BN_mod_exp(ret, A, q, p, ctx))
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| 101 | goto end;
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| 102 | err = 0;
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| 103 | goto vrfy;
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| 104 | }
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| 105 |
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| 106 | if (e == 2) {
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| 107 | /*-
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| 108 | * |p| == 5 (mod 8)
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| 109 | *
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| 110 | * In this case 2 is always a non-square since
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| 111 | * Legendre(2,p) = (-1)^((p^2-1)/8) for any odd prime.
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| 112 | * So if a really is a square, then 2*a is a non-square.
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| 113 | * Thus for
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| 114 | * b := (2*a)^((|p|-5)/8),
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| 115 | * i := (2*a)*b^2
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| 116 | * we have
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| 117 | * i^2 = (2*a)^((1 + (|p|-5)/4)*2)
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| 118 | * = (2*a)^((p-1)/2)
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| 119 | * = -1;
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| 120 | * so if we set
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| 121 | * x := a*b*(i-1),
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| 122 | * then
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| 123 | * x^2 = a^2 * b^2 * (i^2 - 2*i + 1)
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| 124 | * = a^2 * b^2 * (-2*i)
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| 125 | * = a*(-i)*(2*a*b^2)
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| 126 | * = a*(-i)*i
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| 127 | * = a.
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| 128 | *
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| 129 | * (This is due to A.O.L. Atkin,
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| 130 | * Subject: Square Roots and Cognate Matters modulo p=8n+5.
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| 131 | * URL: https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind9211&L=NMBRTHRY&P=4026
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| 132 | * November 1992.)
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| 133 | */
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| 134 |
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| 135 | /* t := 2*a */
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| 136 | if (!BN_mod_lshift1_quick(t, A, p))
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| 137 | goto end;
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| 138 |
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| 139 | /* b := (2*a)^((|p|-5)/8) */
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| 140 | if (!BN_rshift(q, p, 3))
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| 141 | goto end;
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| 142 | q->neg = 0;
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| 143 | if (!BN_mod_exp(b, t, q, p, ctx))
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| 144 | goto end;
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| 145 |
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| 146 | /* y := b^2 */
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| 147 | if (!BN_mod_sqr(y, b, p, ctx))
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| 148 | goto end;
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| 149 |
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| 150 | /* t := (2*a)*b^2 - 1 */
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| 151 | if (!BN_mod_mul(t, t, y, p, ctx))
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| 152 | goto end;
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| 153 | if (!BN_sub_word(t, 1))
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| 154 | goto end;
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| 155 |
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| 156 | /* x = a*b*t */
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| 157 | if (!BN_mod_mul(x, A, b, p, ctx))
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| 158 | goto end;
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| 159 | if (!BN_mod_mul(x, x, t, p, ctx))
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| 160 | goto end;
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| 161 |
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| 162 | if (!BN_copy(ret, x))
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| 163 | goto end;
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| 164 | err = 0;
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| 165 | goto vrfy;
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| 166 | }
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| 167 |
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| 168 | /*
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| 169 | * e > 2, so we really have to use the Tonelli/Shanks algorithm. First,
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| 170 | * find some y that is not a square.
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| 171 | */
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| 172 | if (!BN_copy(q, p))
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| 173 | goto end; /* use 'q' as temp */
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| 174 | q->neg = 0;
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| 175 | i = 2;
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| 176 | do {
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| 177 | /*
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| 178 | * For efficiency, try small numbers first; if this fails, try random
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| 179 | * numbers.
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| 180 | */
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| 181 | if (i < 22) {
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| 182 | if (!BN_set_word(y, i))
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| 183 | goto end;
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| 184 | } else {
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| 185 | if (!BN_priv_rand_ex(y, BN_num_bits(p), 0, 0, 0, ctx))
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| 186 | goto end;
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| 187 | if (BN_ucmp(y, p) >= 0) {
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| 188 | if (!(p->neg ? BN_add : BN_sub) (y, y, p))
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| 189 | goto end;
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| 190 | }
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| 191 | /* now 0 <= y < |p| */
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| 192 | if (BN_is_zero(y))
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| 193 | if (!BN_set_word(y, i))
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| 194 | goto end;
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| 195 | }
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| 196 |
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| 197 | r = BN_kronecker(y, q, ctx); /* here 'q' is |p| */
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| 198 | if (r < -1)
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| 199 | goto end;
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| 200 | if (r == 0) {
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| 201 | /* m divides p */
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| 202 | ERR_raise(ERR_LIB_BN, BN_R_P_IS_NOT_PRIME);
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| 203 | goto end;
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| 204 | }
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| 205 | }
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| 206 | while (r == 1 && ++i < 82);
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| 207 |
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| 208 | if (r != -1) {
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| 209 | /*
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| 210 | * Many rounds and still no non-square -- this is more likely a bug
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| 211 | * than just bad luck. Even if p is not prime, we should have found
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| 212 | * some y such that r == -1.
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| 213 | */
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| 214 | ERR_raise(ERR_LIB_BN, BN_R_TOO_MANY_ITERATIONS);
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| 215 | goto end;
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| 216 | }
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| 217 |
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| 218 | /* Here's our actual 'q': */
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| 219 | if (!BN_rshift(q, q, e))
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| 220 | goto end;
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| 221 |
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| 222 | /*
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| 223 | * Now that we have some non-square, we can find an element of order 2^e
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| 224 | * by computing its q'th power.
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| 225 | */
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| 226 | if (!BN_mod_exp(y, y, q, p, ctx))
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| 227 | goto end;
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| 228 | if (BN_is_one(y)) {
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| 229 | ERR_raise(ERR_LIB_BN, BN_R_P_IS_NOT_PRIME);
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| 230 | goto end;
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| 231 | }
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| 232 |
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| 233 | /*-
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| 234 | * Now we know that (if p is indeed prime) there is an integer
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| 235 | * k, 0 <= k < 2^e, such that
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| 236 | *
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| 237 | * a^q * y^k == 1 (mod p).
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| 238 | *
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| 239 | * As a^q is a square and y is not, k must be even.
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| 240 | * q+1 is even, too, so there is an element
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| 241 | *
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| 242 | * X := a^((q+1)/2) * y^(k/2),
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| 243 | *
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| 244 | * and it satisfies
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| 245 | *
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| 246 | * X^2 = a^q * a * y^k
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| 247 | * = a,
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| 248 | *
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| 249 | * so it is the square root that we are looking for.
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| 250 | */
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| 251 |
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| 252 | /* t := (q-1)/2 (note that q is odd) */
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| 253 | if (!BN_rshift1(t, q))
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| 254 | goto end;
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| 255 |
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| 256 | /* x := a^((q-1)/2) */
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| 257 | if (BN_is_zero(t)) { /* special case: p = 2^e + 1 */
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| 258 | if (!BN_nnmod(t, A, p, ctx))
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| 259 | goto end;
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| 260 | if (BN_is_zero(t)) {
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| 261 | /* special case: a == 0 (mod p) */
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| 262 | BN_zero(ret);
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| 263 | err = 0;
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| 264 | goto end;
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| 265 | } else if (!BN_one(x))
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| 266 | goto end;
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| 267 | } else {
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| 268 | if (!BN_mod_exp(x, A, t, p, ctx))
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| 269 | goto end;
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| 270 | if (BN_is_zero(x)) {
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| 271 | /* special case: a == 0 (mod p) */
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| 272 | BN_zero(ret);
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| 273 | err = 0;
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| 274 | goto end;
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| 275 | }
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| 276 | }
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| 277 |
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| 278 | /* b := a*x^2 (= a^q) */
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| 279 | if (!BN_mod_sqr(b, x, p, ctx))
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| 280 | goto end;
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| 281 | if (!BN_mod_mul(b, b, A, p, ctx))
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| 282 | goto end;
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| 283 |
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| 284 | /* x := a*x (= a^((q+1)/2)) */
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| 285 | if (!BN_mod_mul(x, x, A, p, ctx))
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| 286 | goto end;
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| 287 |
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| 288 | while (1) {
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| 289 | /*-
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| 290 | * Now b is a^q * y^k for some even k (0 <= k < 2^E
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| 291 | * where E refers to the original value of e, which we
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| 292 | * don't keep in a variable), and x is a^((q+1)/2) * y^(k/2).
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| 293 | *
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| 294 | * We have a*b = x^2,
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| 295 | * y^2^(e-1) = -1,
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| 296 | * b^2^(e-1) = 1.
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| 297 | */
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| 298 |
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| 299 | if (BN_is_one(b)) {
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| 300 | if (!BN_copy(ret, x))
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| 301 | goto end;
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| 302 | err = 0;
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| 303 | goto vrfy;
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| 304 | }
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| 305 |
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| 306 | /* find smallest i such that b^(2^i) = 1 */
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| 307 | i = 1;
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| 308 | if (!BN_mod_sqr(t, b, p, ctx))
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| 309 | goto end;
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| 310 | while (!BN_is_one(t)) {
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| 311 | i++;
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| 312 | if (i == e) {
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| 313 | ERR_raise(ERR_LIB_BN, BN_R_NOT_A_SQUARE);
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| 314 | goto end;
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| 315 | }
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| 316 | if (!BN_mod_mul(t, t, t, p, ctx))
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| 317 | goto end;
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| 318 | }
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| 319 |
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| 320 | /* t := y^2^(e - i - 1) */
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| 321 | if (!BN_copy(t, y))
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| 322 | goto end;
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| 323 | for (j = e - i - 1; j > 0; j--) {
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| 324 | if (!BN_mod_sqr(t, t, p, ctx))
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| 325 | goto end;
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| 326 | }
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| 327 | if (!BN_mod_mul(y, t, t, p, ctx))
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| 328 | goto end;
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| 329 | if (!BN_mod_mul(x, x, t, p, ctx))
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| 330 | goto end;
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| 331 | if (!BN_mod_mul(b, b, y, p, ctx))
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| 332 | goto end;
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| 333 | e = i;
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| 334 | }
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| 335 |
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| 336 | vrfy:
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| 337 | if (!err) {
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| 338 | /*
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| 339 | * verify the result -- the input might have been not a square (test
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| 340 | * added in 0.9.8)
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| 341 | */
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| 342 |
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| 343 | if (!BN_mod_sqr(x, ret, p, ctx))
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| 344 | err = 1;
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| 345 |
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| 346 | if (!err && 0 != BN_cmp(x, A)) {
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| 347 | ERR_raise(ERR_LIB_BN, BN_R_NOT_A_SQUARE);
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| 348 | err = 1;
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| 349 | }
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| 350 | }
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| 351 |
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| 352 | end:
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| 353 | if (err) {
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| 354 | if (ret != in)
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| 355 | BN_clear_free(ret);
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| 356 | ret = NULL;
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| 357 | }
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| 358 | if (used_ctx)
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| 359 | BN_CTX_end(ctx);
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| 360 | bn_check_top(ret);
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| 361 | return ret;
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| 362 | }
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